Difference between revisions of "1987 IMO Problems/Problem 5"
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− | {{ | + | Consider the set of points <math>S = \{ (x,x^2) \mid 1 \le x \le n , x \in \mathbb{N} \}</math> in the <math>xy</math>-plane. |
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+ | The distance between any two distinct points <math>(x_1,x^2_1)</math> and <math>(x_2,x^2_2)</math> in <math>S</math> (with <math>x_1 \neq x_2</math>) is: | ||
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+ | <math>d = \sqrt{(x_1-x_2)^2+\left(x^2_1-x^2_2\right)^2} =</math> <math>\sqrt{(x_1-x_2)^2+(x_1-x_2)^2(x_1+x_2)^2} = |x_1-x_2|\sqrt{1+(x_1+x_2)^2}</math>. | ||
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+ | Since <math>1+(x_1+x_2)^2</math> is an integer and not a perfect square, <math>\sqrt{1+(x_1+x_2)^2}</math> is irrational. | ||
+ | Since <math>|x_1-x_2|</math> is a non-zero integer, <math>d</math> is irrational as desired. | ||
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+ | All the points in the set lie on the parabola <math>y = x^2</math>. Thus, it is impossible of any set of three points to be collinear, since no line can intersect a parabola at more than two points. Therefore, any triangle with all vertices in <math>S</math> must be non-degenerate as desired. | ||
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+ | Since all the points are [[lattice points]], by [[Pick's Theorem]], the area of any triangle with all vertices in <math>S</math> must be in the form <math>A = I + \dfrac{B}{2} - 1</math> where <math>I</math> and <math>B</math> are integers. Thus, the area of the triangle must be rational as desired. | ||
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+ | This completes the proof. | ||
{{IMO box|num-b=4|num-a=6|year=1987}} | {{IMO box|num-b=4|num-a=6|year=1987}} | ||
[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] |
Revision as of 00:44, 26 July 2009
Problem
Let be an integer greater than or equal to 3. Prove that there is a set of points in the plane such that the distance between any two points is irrational and each set of three points determines a non-degenerate triangle with rational area.
Solution
Consider the set of points in the -plane.
The distance between any two distinct points and in (with ) is:
.
Since is an integer and not a perfect square, is irrational. Since is a non-zero integer, is irrational as desired.
All the points in the set lie on the parabola . Thus, it is impossible of any set of three points to be collinear, since no line can intersect a parabola at more than two points. Therefore, any triangle with all vertices in must be non-degenerate as desired.
Since all the points are lattice points, by Pick's Theorem, the area of any triangle with all vertices in must be in the form where and are integers. Thus, the area of the triangle must be rational as desired.
This completes the proof.
1987 IMO (Problems) • Resources | ||
Preceded by Problem 4 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 6 |
All IMO Problems and Solutions |